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EULER LINE Re: Midpoint of NO

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  • Francisco Javier
    More specifically: Let ABC be a triangle, P a point and k a real number. Call A , B , C ,P the points on lines AP, BP, CP such that
    Message 1 of 3 , Mar 28, 2013
      More specifically:

      Let ABC be a triangle, P a point and k a real number.

      Call A', B', C',P' the points on lines AP, BP, CP such that AA':A'P=BB':B'P=CC':C'P=OO':O'P=k:1.

      Then the radical center R of circles (A, AA'), (B, BB'), (C, CC') is the inverse of P with respect the circle (O, OO').




      --- In Hyacinthos@yahoogroups.com, "Francisco Javier" <garciacapitan@...> wrote:
      >
      > For a general P instead of H, your locus is the line OP, thus for P=H we have the Euler line.
      >
      > Francisco Javier.
      >
      > --- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis <anopolis72@> wrote:
      > >
      > > Let ABC be a triangle and A*, B*, C* three points on AH,BH,CH, resp,
      > > such that:
      > >
      > > AA* / AH = BB* / BH = CC* / CH = t
      > >
      > > Which is the locus of the radical center of the circles
      > > (A, AA*), (B, BB*), (C,CC*) ?
      > >
      > > APH
      > >
      > > On Thu, Mar 28, 2013 at 2:04 AM, Antreas <anopolis72@> wrote:
      > >
      > > > **
      > > >
      > > >
      > > > Let ABC be a triangle and A',B',C' the midpoints
      > > > of AH,BH,CH, resp. (A'B'C' = Euler triangle)
      > > >
      > > > The radical center of the circles (A,AA'), (B,BB'), (C,CC')
      > > > is the midpoint of NO
      > > >
      > > > aph
      > > >
      > > >
      > > >
      > > >
      > >
      > >
      > > [Non-text portions of this message have been removed]
      > >
      >
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